Answer
$\dfrac{2}{\sqrt \pi } e^{-3t} \dfrac{1}{\sqrt t}$
Work Step by Step
Since, $F(s)=\dfrac{2}{\sqrt {s+3}}$
The inverse Laplace transform of function can be expressed as:
$F(t)=\dfrac{1}{\sqrt t} $
This yields: $F(s)=\int_0^{\infty} \dfrac{e^{-st}}{\sqrt t} \ dt=\sqrt s\int_0^{\infty} \dfrac{e^{-st}}{\sqrt {st}} \ dt$
Now, apply the first shifting Theorem.
$F(s)=\sqrt {\dfrac{\pi}{s}}$
So, $f(t)=\dfrac{2}{\sqrt \pi } e^{-3t} \dfrac{1}{\sqrt t}$
